Question: $h(x)=3x^2-2x$ What is the average rate of change of $h$ over the interval $-3\leq x \leq t$, in terms of $t$, where $t\neq -3$ ? Your answer must be fully expanded and simplified.
Answer: This is the formula for the average rate of change of a function $f$ over the interval $[a,b]$ : $\dfrac{f(b)-f(a)}{b-a}$ We can calculate that $h(-3)=33$. We are interested in the average rate of change of $h(x)=3x^2-2x$ over the interval $-3\leq x \leq t$ : $\begin{aligned} &\phantom{=}\dfrac{h(t)-h(-3)}{(t)-(-3)} \\\\ &=\dfrac{3t^2-2t-33}{t-(-3)} \\\\ &=\dfrac{3t^2-2t-33}{t+3} \end{aligned}$ We can now simplify the expression we obtained. $\begin{aligned} \dfrac{3t^2-2t-33}{t+3}&=\dfrac{(3t-11)(t+3)}{t+3} \\\\ &=3t-11\text{, for }t\neq -3 \end{aligned}$ Since we are given that $t\neq -3$, the average rate of change of the function is $3t-11$. Notice that the average rate of change is calculated just like the slope of the secant line that intersects the graph of the function at the interval's endpoints.